Question

A gun has a muzzle speed of 80 meters per second. What angle of elevation should be used to hit an object 180 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravity.
The answer must be in radians.

I found: angle=1/2arcsin (180*9.8)/(80^2), but it was not correct.

Answer 1

Hi. Your working equation is correct. The general equation to be used is 

$Range = sin(2 \alpha )( \frac{ v^{2} }{g} )$

However, make sure that when you input this in the calculator, it is in the mode of rad. 

The answer is 0.14 rad.

I hope I was able to answer your question. Have a good day.

Answer 2

Knowing the initial velocity and angle, the horizontal range formula is given by   R= V^2sin(2teta) / g, so we can get 
sin(2teta)=Rg/V^2
sin(2teta)= (180 x 9.8)/ 80^2= 0.27, sin(2teta)=0.27, 2teta=arcsin(0.27)=15.66, so teta=15.66/2

teta=7.83°